Let $F$ be a closed subspace of a normed space. How to prove that $E$ is complete iff $F$ and $E/F$ are both complete ?
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1Use the fact that a space is complete iff absolutely convergent series also converge. Then show that $E/F$ is Banach. Since $F$ is closed, and $E$ complete, a cauchy sequence in $F$ converges to some $x$, but since $F$ is closed, it converges in $F$. – Andres Mejia Sep 22 '16 at 17:13
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1Regarding the second comment, see the more general statement: http://math.stackexchange.com/questions/240640/a-closed-subspace-of-a-banach-space-is-a-banach-space – Andres Mejia Sep 22 '16 at 17:13
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2It is not duplicate. The question which you gave the link to proves one side of the claim ($X$ complete and $Y$ closed $\Rightarrow$ $X/Y$ complete). The question here asks to prove the reverse implication also( which is not there in the given link). – learning_math Sep 23 '16 at 14:39
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Let $\pi: E \to E/F$ be the quotient map. Let $x_n$ be a Cauchy sequence in $E$. Then $\pi(x_n)$ is Cauchy in $E/F$, so has a limit which is $\pi(z)$ for some $z \in E$. Since $\|\pi(x_n - z)\| = \inf \{\|x_n - z - y\|: y \in F\}$, there is $y_n \in F$ with $\|x_n - z - y_n\| < \|\pi(x_n - z) \| + 1/n$, and thus $\|x_n - z - y_n\| \to 0$. Show that $y_n$ is Cauchy, and then that $x_n$ converges to $z + \lim_n y_n$.
Robert Israel
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Sorry, $\pi_n$ was a typo (fixed). $y_n$ is Cauchy because $|y_n - y_m| \le |x_n - z - y_n| +|x_m - x_n| + |x_m - z - y_m| $. – Robert Israel Sep 22 '16 at 20:00
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@RobertIsrael What if we take $E=F$ and $E$ not complete? where this proof use the fact that $F$ is proper subspace. – Meet Patel Jan 24 '24 at 14:06
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1@MeetPatel If $E = F$, $E/E = {(0)}$ is complete, so the statement just says $E$ is complete iff $E$ is complete. – Robert Israel Jan 24 '24 at 15:14
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